What is the distance formula between two points on a number line?

The distance d between two points a and b on a number line is given by d = |a - b| = |b - a|. Because of the absolute value, the distance is always positive or zero.

What is the distance formula between two points on a plane?

Given two points A = (ax, ay) and B = (bx, by) on a coordinate plane, the distance d between them is given by d = √((bx - ax)^2 + (by - ay)^2), derived using the Pythagorean theorem.

How do you find the midpoint of a line segment?

The midpoint M of a line segment with endpoints A = (ax, ay) and B = (bx, by) is found by taking the average of their x-coordinates and y-coordinates: M = ((ax + bx)/2, (ay + by)/2).

Distance and Midpoint Formulas | Step-by-Step Geometry Solutions

The distance between two points and the midpoint of a segment formulas are presented along with examples, practice questions, and detailed solutions.

A distance and midpoint calculator is also available to check your answers.

Distance on a Number Line & Pythagorean Theorem

Let us first define the distance \( d \) between two points \( A \) and \( B \) whose coordinates are respectively \( a \) and \( b \) on a number line as:

\[ d = |a - b| = |b - a| \]

Fig.1 - Distance Between Points on the Number Line

Note: Because of the absolute value, the distance between points is always positive or equal to zero.

We now review the Pythagorean theorem: given a right triangle, as shown in figure 2 below, the hypotenuse (side opposite the right angle) and the two sides are related by:

\[ h^2 = x^2 + y^2 \quad \text{or} \quad h = \sqrt{x^2 + y^2} \]

Distance Between Two Points on a Plane

We now use the distance on a number line and the Pythagorean theorem to write the formula for the distance between any two points on a plane. Given two points \( A = (a_x, a_y) \) and \( B = (b_x, b_y) \), the distance \( d \) between them (the length of segment \( AB \)) is given by:

\[ \large \color{red}{d = \sqrt{(b_x - a_x)^2 + (b_y - a_y)^2} \qquad (I)} \]

Fig.3 - Distance Between two Points on a Plane

Note: \( |b_x - a_x|^2 = (b_x - a_x)^2 \) and \( |b_y - a_y|^2 = (b_y - a_y)^2 \)

Example: Find the distance \( d \) between \( A = (3,2) \) and \( B = (-3,-6) \)

According to formula (I), the distance \( d \) is given by:

\[ d = \sqrt{(-3 - 3)^2 + (-6 - 2)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

Midpoint Definition and Formula

The midpoint \( M \) of the line segment \( AB \) is the point on the segment such that the lengths of segments \( MA \) and \( MB \) are equal (see Figure 4).

\[ \large \color{red}{M = \left(\frac{a_x + b_x}{2} \; , \; \frac{a_y + b_y}{2}\right)} \qquad (II) \]

Note: The \( x \) and \( y \) coordinates of \( M \) are the averages of the \( x \) and \( y \) coordinates of points \( A \) and \( B \), respectively.

Example: Find the midpoint of segment \( AB \) given \( A = (5,2) \) and \( B = (-7,-10) \)

According to formula (II), the coordinates of midpoint \( M \) are:

\[ M = \left(\frac{5 + (-7)}{2} , \frac{2 + (-10)}{2}\right) = \left(\frac{-2}{2} , \frac{-8}{2}\right) = (-1 , -4) \]

Practice Questions and Detailed Solutions

Part A: Distance on a Number Line

Find the distance between each pair of points in the number lines shown below.

The coordinate of point \( A \) is \( -2 \) and point \( B \) is \( 9 \): \[ d_1 = |-2 - 9| = |-11| = 11 \quad \text{or} \quad d_1 = |9 - (-2)| = |11| = 11 \]
The coordinate of point \( C \) is \( -8 \) and point \( D \) is \( -3 \): \[ d_2 = |-8 - (-3)| = |-8 + 3| = |-5| = 5 \quad \text{or} \quad d_2 = |-3 - (-8)| = |5| = 5 \]
The coordinate of point \( E \) is \( 3 \) and point \( F \) is \( 10 \): \[ d_3 = |3 - 10| = |-7| = 7 \quad \text{or} \quad d_3 = |10 - 3| = |7| = 7 \]

Part B: Distance Between Two Points on a Plane

Find the distance between the given pair of points:

\( A = (0,0) , B = (-4,3) \) \[ d(AB) = \sqrt{(-4 - 0)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
\( C = (-2,-2) , D = (6,4) \) \[ d(CD) = \sqrt{(6 - (-2))^2 + (4 - (-2))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]
\( E = (4,7) , F = (-4,7) \) \[ d(EF) = \sqrt{(-4 - 4)^2 + (7 - 7)^2} = \sqrt{(-8)^2 + 0} = \sqrt{64} = 8 \]

Part C: Midpoint Between Segments

Find the midpoint between the segments defined by each pair of points:

\( A = (-8,0) , B = (-4,4) \) \[ M_{AB} = \left(\frac{-8 + (-4)}{2} , \frac{0 + 4}{2}\right) = \left(\frac{-12}{2} , \frac{4}{2}\right) = (-6, 2) \]
\( C = (-3,-4) , D = (8,3) \) \[ M_{CD} = \left(\frac{-3 + 8}{2} , \frac{-4 + 3}{2}\right) = \left(\frac{5}{2} , -\frac{1}{2}\right) \]
\( E = (4,7) , F = (-4,7) \) \[ M_{EF} = \left(\frac{4 + (-4)}{2} , \frac{7 + 7}{2}\right) = (0, 7) \]

Part D: Finding an Endpoint Given Midpoint

Given points \( A = (x_0, y_0) \) and \( B = (-4,8) \). Find the coordinates of point \( A \) if the midpoint of segment \( AB \) is \( M = (-2,0) \).

Using the midpoint formula: \( M = \left(\frac{x_0 - 4}{2} , \frac{y_0 + 8}{2}\right) = (-2, 0) \)

Setting up equations for coordinates:

\[ \frac{x_0 - 4}{2} = -2 \implies x_0 - 4 = -4 \implies x_0 = 0 \] \[ \frac{y_0 + 8}{2} = 0 \implies y_0 + 8 = 0 \implies y_0 = -8 \]

Therefore, \( A = (0, -8) \).

Part E: Finding Variables in Midpoint Equations

Given \( A = (3x_0, -y_0) \) and \( B = (x_0, 4y_0) \). Find \( A \) and \( B \) if the midpoint is \( M = (2,6) \).

Using the midpoint formula:

\[ M = \left(\frac{3x_0 + x_0}{2} , \frac{-y_0 + 4y_0}{2}\right) = \left(\frac{4x_0}{2} , \frac{3y_0}{2}\right) = (2, 6) \]

Solving for \( x_0 \) and \( y_0 \):

\[ \frac{4x_0}{2} = 2 \implies 4x_0 = 4 \implies x_0 = 1 \] \[ \frac{3y_0}{2} = 6 \implies 3y_0 = 12 \implies y_0 = 4 \]

Substituting back: \( A = (3(1), -4) = (3,-4) \) and \( B = (1, 4(4)) = (1,16) \).

Part F: Comprehensive Segment Analysis

Find midpoint \( M \) of \( A = (-4,1) \) and \( B = (2,5) \): \[ M = \left(\frac{-4 + 2}{2} , \frac{1 + 5}{2}\right) = (-1, 3) \]
Find lengths of \( MA \) and \( MB \): \[ d(MA) = \sqrt{(-4 - (-1))^2 + (1 - 3)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \] \[ d(MB) = \sqrt{(2 - (-1))^2 + (5 - 3)^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \]
Yes, they are equal as expected from the definition of the midpoint.
Find length of segment \( AB \): \[ d(AB) = \sqrt{(2 - (-4))^2 + (5 - 1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \]
Show \( d(MA) \) is half of \( d(AB) \): \[ \frac{1}{2} \times d(AB) = \frac{1}{2} \times 2\sqrt{13} = \sqrt{13} = d(MA) \]